Answer:

Correct Answer: D

Solution:

The general term of the series is $ =\frac{2*2^{r}-1}{2^{r}} $

we can write the given sum as $ 1+\sum\limits _{r=1}^{19}{\frac{2*2^{r}-1}{2^{r}}} $

Now, $ \sum\limits _{r=1}^{19}{\frac{2*2^{r}-1}{2^{r}}=\sum\limits _{r=1}^{19}{2-\frac{1}{2^{r}}=2(19)-}} $

$ \frac{\frac{1}{2}(1-{{(\frac{1}{2})}^{19}})}{1-\frac{1}{2}}=38+\frac{{{(\frac{1}{2})}^{19}}-1}{1}=38+{{(\frac{1}{2})}^{19}}-1=37+{{(\frac{1}{2})}^{19}} $ As we can write the given sum as $ 1+\sum\limits _{r=1}^{19}{\frac{2*2^{r}-1}{2^{r}}} $

$ \Rightarrow $ $ 1+37+{{(\frac{1}{2})}^{19}}=38+{{(\frac{1}{2})}^{19}} $

Therefore correct answer is D $ T _{n}=\frac{2^{n}-1}{{2^{n-1}}} $ $ =2-\frac{1}{{2^{n-1}}} $ $ \sum{_0^{20}T _{n}=\sum{_0^{20}2-\sum{_0^{20}}}\frac{1}{{2^{n-1}}}} $

$ =2(20)-\frac{(1)( 1-\frac{1}{2^{20}} )}{1-\frac{1}{2}} $ $ =40-2( 1-\frac{1}{2^{20}} ) $ $ =38+\frac{1}{2^{19}} $ Hence, the answer is option D Let $ S _{n}=1+\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+….+T _{n} $ Divide each term by 2,

$ \Rightarrow $ $ \frac{S _{n}}{2}=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+….+{T _{n+1}} $ ..(i)

$ \Rightarrow $ $ \frac{S _{n}}{2}=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+…+T _{n}+{T _{n+1}} $

$ \Rightarrow $ $ 0=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+…+\frac{1}{2^{n}}-{T _{n+1}} $

$ \Rightarrow $ $ {T _{n+1}}=\frac{1}{2}+\frac{1}{4}+….+\frac{1}{2^{n}} $ This is a sum of n terms of a G.P. with common ratio $ \frac{1}{2}<1 $ and first term $ \frac{1}{2} $

$ \Rightarrow $ $ {T _{n+1}}=\frac{\frac{1}{2}(1-{{(\frac{1}{2})}^{n}})}{1-\frac{1}{2}} $

$ \Rightarrow $ $ {T _{n+1}}=\frac{2^{n}-1}{2^{n}} $ Now, $ \frac{S _{n}}{2}=\sum{\frac{2^{n}-1}{2^{n}}} $

$ \Rightarrow $ $ \frac{S _{20}}{2}=\sum\limits _{n=1}^{20}{\frac{2^{n}-1}{2^{n}}} $

$ \Rightarrow $ $ \frac{S _{20}}{2}=\sum\limits _{n=1}^{20}{1-\frac{1}{2^{n}}} $

$ \Rightarrow $ $ \frac{S _{20}}{2}=\sum\limits _{n=1}^{20}{1-\sum\limits _{n=1}^{20}{\frac{1}{2^{n}}}} $

$ \Rightarrow $ $ \frac{S _{20}}{2}=20-( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+…+\frac{1}{20} ) $

$ \Rightarrow $ $ \frac{S _{20}}{2}=20-( \frac{\frac{1}{2}( 1-( \frac{1}{2} ) )}{1-\frac{1}{2}} ) $

$ \Rightarrow $ $ \frac{S _{20}}{2}=20-1+\frac{1}{2^{20}} $

$ \Rightarrow $ $ S _{20}=38+\frac{1}{2^{19}} $